Can someone helping me!

For this activity create a kinship chart of a celebrity family or fictional family on television or in other media. Choose a person as “ego” and use the anthropological symbols to identify as many of their relatives as you can. Be sure to choose a family with at least three known generations.

Answer:

2025 J

Explanation:

The work done by gravity on Logan can be calculated using the formula W = mgh, where W is the work done by gravity, m is the mass of Logan, g is the acceleration due to gravity and h is the change in height.

First we need to calculate the change in height. The initial height of Logan can be calculated using trigonometry. The vertical component of the rope length when it makes a 30 degree angle with the vertical is 4.10m * cos(30) = 3.55m.

Similarly, when Logan releases the rope at an angle of 13.1 degrees with the perpendicular (or 90-13.1=76.9 degrees with the vertical), his height above water level will be 4.10m * cos(76.9) = 0.93m.

So his change in height will be 3.55m - 0.93m = 2.62m.

Now we can calculate how much work gravity does on him: W = mgh = (79kg)(9.8 m/s^2)(2.62m) ≈ 2025 J.

So gravity does about 2025 J of work on Logan up to the point where he releases the rope.

Answer:

46.6 m/s in the positive y-direction.

Explanation:

Self-care can be demonstrated in a healthy way by recalling a happy memory.

The first step toward self-care is allowing yourself to be honest and vulnerable. Through self-reflection, you can at last let out all that you've been keeping down. Your journal is your very own private platform on which you can record your successes, anxieties, and worries.

Taking care of oneself exercises can go from proactive tasks, for example, practicing and practicing good eating habits, to mental exercises like perusing a book or rehearsing care, to otherworldly or social exercises, for example, imploring or getting lunch with a companion.

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The location at the point where electric field becomes zero is at 3.6m, when a positive point charge Q1= 2.6×10−5 C is fixed at the origin and a negative point charge Q2= −5.1×10−6 C is fixed at 2.0m.

Electric field is field around electrically charged particle where columbic force of attraction or repulsion can be experienced by other charged particles. It is denoted by letter E and it's SI unit is V/m Volt per meter or N/C newton per coulomb.

Electric field comes inward to the center of the negative charge and it is going outward for positive charge.

Given,

Q₁= 2.6×10⁻⁵ C

Q₂= −5.1×10⁻⁶ C

x = 2.0m,

In this case there are two charges, one at the origin and other is at 2m distance from origin. Both these charges are kept along positive x axis. looking at the situation there are 2 cases

Case 1 ( electric field in between two charges)

in this case direction of the resultant electric field in between two charges is along positive x direction (direction of electric field of positive charge is outward along x axis which is along positive x direction and direction of electric field of negative charge is inward along x axis which is also in positive x direction).

Case 2 ( right side of the negative charge)

In this case direction of electric field of positive charge is outward along x axis which is along positive x direction and direction of electric field of negative charge is inward along x axis but which is in opposite direction of electric field due to positive charge.

Hence our electric field is zero at right side of the negative charge.

on right side, electric fields due to both charges are opposite to each other then we can write mathematically E₁ - E₂ = 0 at the point where resultant electric field is zero. Where E₁ is elctric field due to positive charge and E₂ is electric field due to negative charge.

E₁ - E₂ = 0

E₁ = E₂

[tex]\frac{kQ_{1}}{x^{2}} = \frac{kQ_{2} }{(x+2)^{2}}[/tex]

[tex]\frac{Q_{1}}{x^{2}} = \frac{Q_{2} }{(x+2)^{2}}[/tex]

[tex]\frac{Q_{1}}{Q_{2} }} = \frac{x^{2} }{(x+2)^{2}}[/tex]

(2.6×10⁻⁵ C ÷ -5.1×10⁻⁶ C)  [tex]= \frac{x^{2} }{(x+2)^{2}}[/tex]

[tex]\frac{x^{2} }{(x+2)^{2}}[/tex] = 5.098 ≅ -5.1

x² = 5.1 (x+2)²

x = 2.25(x+2)

x=2.25x + 4.5

x(1-2.25) = 4.5

x(-1.25) = 4.5

x = 3.6m

Hence Electric field is zero at 3.6m from origin.

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Two points in the xy plane have Cartesian coordinates (5.00, −5.50) m and (−7.00, 7.00) m.So,  the distance between two points in the xy-plane is 17.32 m.

What is Distance?

Distance is a measurement in numbers of the area between two locations or objects. It is a scalar quantity that symbolises the distance between two places in space or the length of a path taken by an object. Depending on the situation and the scale of the items being measured, distance can be expressed in a variety of ways, including metres, kilometres, miles, feet, and so forth.

Displacement, a vector number that denotes the shift in an object's location with respect to a reference point, is frequently distinguished from distance in physics. Displacement, as opposed to distance, which only analyses the size of the path travelled, considers both the distance travelled by an object and its direction of motion.

To find the distance between two points in the xy-plane, we use the distance formula:

d = [tex]\sqrt{(x2 - x1)^2 + (y2 - y1)^2}[/tex]

where (x1, y1) and (x2, y2) are the coordinates of the two points.

Using the given coordinates, we have:

x1 = 5.00 m, y1 = -5.50 m

x2 = -7.00 m, y2 = 7.00 m

Substituting these values into the formula, we get:

d = [tex]\sqrt{(-7.00 - 5.00)^2 + (7.00 - (-5.50))^2}[/tex]

d= [tex]\sqrt{(-12.00)^2 + (12.50)^2}[/tex]

d= [tex]\sqrt{(144.00 + 156.25)}[/tex]

d= [tex]\sqrt{(300.25)}[/tex]

d=17.32 m

Therefore, the distance between the two points is approximately:

d = 17.32 m (rounded to two decimal places)

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Answer:

To calculate the work being done by Sally, we need to use the formula:

Work = Force x Distance x cos(theta)

where:

Force = 10 N (the force required to lift the paintbrush)

Distance = 10 m (the distance Sally lifts the paintbrush)

theta = 0 degrees (since Sally is lifting the paintbrush straight up, the angle between the force and the direction of motion is 0 degrees)

So, the work done by Sally is:

Work = 10 N x 10 m x cos(0)

Work = 100 joules

Therefore, the joules of work being put out by Sally when she lifts the 10 N paintbrush at a distance of 10 meters is 100 joules.

Explanation:

Answer:

1 Minutes =2Pi/60

21 Minutes=2Pi/60*21

so the answer is, 2.198

Explanation:

we know that

there is 60 min one hour

so the there is 2Pi radians

so 21 minutes is 21/60

and we know the clock hand will travel anticlock wise

and a full circle measures 2Pi radians

and we know that one minute =2Pi /60 radians

so, 21 minutes = (21Pi/60)*60

(Pi=3.14)

21 minutes = 2.19

The initial rotational kinetic energy of the yo-yo is equal to ½ × 3.5 ×10-5 kg×m2 ×(0.70m/ 0.008m)2 = 0.000912 Joules.

Kinetic energy is the energy of motion. It is the energy an object has due to its motion. Kinetic energy can be calculated by multiplying the mass of an object with the square of its velocity. Kinetic energy is a form of energy that is associated with the movement of objects. It is released when objects collide or when objects are deformed. Kinetic energy can also be converted into other forms of energy, such as thermal energy, sound energy and electrical energy.

The initial gravitational potential energy of the yo-yo is equal to the change in height multiplied by the weight of the yo-yo. The weight of the yo-yo can be calculated using the equation W = m × g, where m is the mass and g is the acceleration due to gravity. Therefore, the initial gravitational potential energy of the yo-yo is equal to 0.05 kg ×9.8 m/s2 ×0.70 m = 0.343 Joules.

b. Calculate the initial rotational kinetic energy of the yo-yo.

The initial rotational kinetic energy of the yo-yo is equal to ½Iω2, where I is the rotational inertia and ω is the angular velocity. The angular velocity can be calculated using the equation ω = v/r, where v is the linear velocity and r is the radius of the yo-yo. Therefore, the initial rotational kinetic energy of the yo-yo is equal to ½ × 3.5 ×10-5 kg×m2 ×(0.70m/ 0.008m)2 = 0.000912 Joules.

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The 35m would the Roche limit be for an earth-orbiting body with the same density as iron(8.0g/cm ^3).

What is density ?

How much material an object contains per unit volume is determined by the density of the substance. It is symbolized by the letter D, however it can also be represented by the symbol.  The density of a substance reveals how dense it is in a certain space. The definition of density is the mass of a substance per unit volume.

It is measured in meters per second in the SI (ms -1 ). A body is considered to be accelerating if there is a change in the magnitude or direction of its velocity.

Therefore, The 35m would the Roche limit be for an earth-orbiting body with the same density as iron(8.0g/cm ^3).

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Answer:

h(t) = -16t^2 + 1000

Explanation:

The quadratic function that represents the motion of a freely falling ballast bag that starts from a resting position on a balloon 1000 feet above the ground is given by the equation h(t) = -16t^2 + 1000, where h(t) is the height of the ballast bag in feet at time t seconds after it is released, and -16t^2 represents the effect of gravity on the ballast bag as it falls.

Answer:

-16t^2 + 1000

Explanation:

The kinetic energy (KE) of an object is given by the formula:

KE = (1/2) * m * v^2

where m is the mass of the object and v is its velocity.

Plugging in the given values, we get:

KE = (1/2) * 60 kg * (20 m/s)^2

KE = 12,000 joules

Therefore, the kinetic energy of the car is 12,000 joules.

Answer:

Since the net torque is negative, this means that the counterclockwise torque is greater than the clockwise torque. As a result, the wheel will rotate in the counterclockwise direction. Therefore, the answer is:

The wheel will rotate counterclockwise.

Explanation:

To determine the resulting motion of the wheel, we need to calculate the net torque acting on the wheel. Clockwise torque is positive, while counterclockwise torque is negative. Therefore, the net torque can be calculated as follows:

Net torque = (Clockwise torque) - (Counterclockwise torque)

Net torque = (24.4 N) x (0.22 m) - (7.9 N) x (0.74 m)

Net torque = 5.368 Nm - 5.846 Nm

Net torque = -0.478 Nm

Kinship charts, often known as kinship diagrams, show relationships. Similar to a family tree chart or a pedigree chart, you can use a kinship diagram to display your genealogy.

An illustration of relationships in a family, community, or culture is called a kinship diagram. Kinship charts resemble family trees in many ways. Therefore, kinship diagrams are used more broadly to comprehend how most families in a culture function rather than identifying individual names or modeling the diagram after one family.

Kinship diagrams enable cultural anthropologists to swiftly sketch out relationships between people throughout the interview process. Also, it offers a way to represent kinship patterns visually in cultures without the use of names, which can be confusing, and it protects people's privacy.

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The gravitational potential energy of the object is -176J. Option 4 is the answer.

The gravitational potential energy (U) of an object of mass (m) at a height (h) relative to a reference point where gravitational potential energy is defined to be zero is given by:

U = mgh

where g is the acceleration due to gravity (9.81 m/s^2).

In this problem, the object is 3.00 m below the ground relative to a point 5.00 m below the ground. So, the height of the object relative to the reference point is:

h = -3.00 m - (-5.00 m) = 2.00 m

Note that we use a negative sign for h because the object is below the reference point.

Plugging in the values, we get:

U = (9.00 kg) x (9.81 m/s^2) x (-2.00 m) = -176.22 J

Rounding off to two significant figures, the gravitational potential energy of the object is -176 J. Therefore, the answer is -176J.

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Answer & Explanation:

Friction is useful in the following situations:

The brakes on a bike: Friction is necessary in the brakes of a bike as it allows the brake pads to grip onto the rim or disc, which helps to slow down or stop the bike.

Inside the hub of the wheel on a bike: Friction is necessary to prevent the wheel from spinning too freely, which could lead to loss of control or accidents. The hub's bearings require just the right amount of friction to keep the wheel spinning freely but not too fast.

Friction is a nuisance in the following situation:

The moving parts of the chain and gears: Friction between the moving parts of the chain and gears can cause the bike to lose efficiency and require more effort from the rider. This can cause wear and tear on the bike and make it harder to ride.

The angle of application of the force ​ is theta = arccos(500 J / (F * 20 m))

We can use the work-energy principle to solve this problem. The work done by the gardener's force is equal to the change in kinetic energy of the mower. Since the mower starts from rest, the initial kinetic energy is zero. Therefore, the work done by the gardener's force is equal to the final kinetic energy of the mower:

Work = Change in kinetic energy

500 J = (1/2)mv^2

where m is the mass of the mower and v is its final velocity.

Since we don't know the mass of the mower or its final velocity, we cannot directly solve for the angle of application of the force. However, we can use the fact that the work done by a force is equal to the product of the magnitude of the force, the displacement, and the cosine of the angle between the force and the displacement:

Work = Fd cos(theta)

where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and the displacement.

Substituting the given values, we have:

500 J = F(20 m) cos(theta)

Solving for the cosine of the angle:

cos(theta) = 500 J / (F * 20 m)

Since we don't know the magnitude of the force, we cannot directly solve for the angle. However, we can rearrange the equation to isolate the angle:

theta = arccos(500 J / (F * 20 m))

Therefore,  we need to know the magnitude of the force to find the angle of application.

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The terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.

How to calculate terminal velocity?

To calculate the terminal velocity of the block, we need to use the equation for terminal velocity which is:

v_terminal = sqrt((2mg)/(ρAC_d))

where:

First, let's convert the dimensions of the block from centimeters to meters:

width = 16.0 cm = 0.16 m

length = 19.0 cm = 0.19 m

height = 25.0 cm = 0.25 m

Next, let's calculate the cross-sectional area of the block:

A = width * length = 0.16 m * 0.19 m = 0.0304 m^2

Now we can substitute the values into the equation for terminal velocity:

v_terminal = sqrt((24.55 kg9.81 m/s^2)/(1.43 kg/m^30.0304 m^20.822))

v_terminal = 2.08 m/s

Therefore, the terminal velocity of the block is approximately 2.08 m/s when it is traveling downward with a drag coefficient of 0.822.

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Answer:

The moon does not have an atmosphere because it does not have enough gravity to hold onto the gases that make up an atmosphere. The moon also does not have plate tectonic activity because it does not have enough internal heat to drive the movement of the plates.

Explanation:

Mission control is an exciting career that involves managing space missions from the ground.

Mission control is an exciting career that involves managing space missions from the ground. I chose this career because of my interest in space exploration and the opportunity to contribute to the advancement of space technology.

Some questions that could be asked about this career include:

What qualifications are required to become a mission control specialist?

What kind of training do mission control specialists undergo?

What are the typical responsibilities of a mission control specialist during a space mission?

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The change in momentum of the ball is 1.0 kg m/s [E].

The change in momentum of the ball can be calculated using the formula:

Δp = p₂ - p₁

where Δp is the change in momentum, p₂ is the final momentum, and p₁ is the initial momentum.

The momentum of an object is defined as the product of its mass and velocity, so we can calculate the initial and final momenta of the ball as follows:

p₁ = m₁v₁ = (0.1 kg)(5 m/s [W]) = -0.5 kg m/s

p₂ = m₁v₂ = (0.1 kg)(5 m/s [E]) = 0.5 kg m/s

Substituting these values into the formula for Δp, we get:

Δp = p₂ - p₁ = (0.5 kg m/s) - (-0.5 kg m/s) = 1.0 kg m/s [E]

Therefore, the change in momentum of the ball is 1.0 kg m/s [E] using Δp = p₂ - p₁.

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Answer:

475 J

Explanation:

At the top of the feature, Robert has 475 J of kinetic energy and 2240 J of gravitational potential energy. According to the law of conservation of energy, the total energy of a system is constant, so the sum of the kinetic energy and gravitational potential energy at the top of the feature is equal to the sum of the kinetic energy and gravitational potential energy at the bottom of the feature.

Since Robert is at the bottom of the feature, he has lost all of his gravitational potential energy. Therefore, the total energy he has at the bottom of the feature is equal to his kinetic energy at the top of the feature:

Total energy = Kinetic energy at top = 475 J

Therefore, the answer is 475 J, which is option D.

(a) Time taken, T is 11.83 s

(b) Acceleration is 1.428 m/s²

A body is said to be in uniform acceleration when it is moving in a straight line with an increase in velocity in equal intervals of time.

Here,

s = 100 m

v = u + at

a = (v - u)/t = (10 - 0)/7

a = 10/7 m/s²

(a) s = ut + 1/2 at²

   100 = 0 + 1/2 x 10/7 x t²

    t² = 140

    t = √140 = 11.83 s

(b) Acceleration, a = 10/7 =1.428 m/s²

Hence,

(a) Time taken, T is 11.83 s

(b) Acceleration is 1.428 m/s²

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Sound waves can travel through a medium (such as air, water, or solid objects) and can reflect or diffract around obstacles, allowing them to be heard even when there is no direct path to the source of the sound.

When sound waves encounter an obstacle, they can either reflect or diffract. Reflection occurs when sound waves bounce off a surface and return in the direction they came from. Diffracton occurs when sound waves bend around an obstacle and continue in a different direction. This allows sound waves to travel around corners and through small openings.

For example, if someone is talking behind a wall, the sound waves they produce can diffract around the wall and reach your ears, allowing you to hear them. Similarly, when you hear an echo, it is because sound waves have reflected off a surface and reached your ears.

In some cases, sound waves can even travel through solid objects. This is known as vibration or bone conduction. For example, when you hear your own voice or a piece of music through a set of headphones, the sound waves are transmitted through the headphones and vibrate the bones in your ear, which sends signals to your brain that you perceive as sound.

In summary, sound waves can travel through a medium and reflect or diffract around obstacles, allowing them to be heard even when there is no direct path to the source of the sound.

Answer: 900 m

Explanation: If you walk at a speed of 5 meters per second for 3 minutes, or 180 seconds, then you will have walked 900 meters. This can be seen in the equation [tex]x=x_{0}+v_{x0}t+\frac{1}{2} a_{x}t^2[/tex]. As the velocity is constant (no acceleration) and the initial position is zero, plugging in known values for v and t, you can solve for x of 900 m.

The bike is going at a speed of 7.76 m/s (or about 28 km/h) while you are pushing it with a force of 190 N over a distance of 2.0 m.

We can use the work-energy principle to determine the final velocity of the bike. The work-energy principle states that the net work done on an object is equal to its change in kinetic energy. In this case, the net work is done by the force that you apply to the bike over a distance of 2.0 m:

Net work = force x distance x cos(theta) = 190 N x 2.0 m x cos(0) = 380 J

The change in kinetic energy of the bike is given by:

[tex]Delta K = 1/2 mv^2 - 0[/tex]

where m is the mass of the bike and v is its final velocity.

Setting the net work equal to the change in kinetic energy, we have:

[tex]380 J = 1/2 (22 kg) v^2[/tex]

Solving for v, we get:

[tex]v = sqrt(2 x 380 J / 22 kg) = 7.76 m/s[/tex]

Therefore, the bike is going at a speed of 7.76 m/s (or about 28 km/h) while you are pushing it with a force of 190 N over a distance of 2.0 m.

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If object 1 and 2 attracts each other with an electrostatic force of 18N and is separated by some distance, Then if that distance will be halved then the electrostatic force will be = 18×4=72N.

It states that" The force between two charges is directly proportional to the product of their magnitude and inversely proportional to the square of the distance between the charges and acts along the line joining the two charges."

Mathematically,

F∝q1×q2

F∝1/[tex]r^{2}[/tex]

i.e

F= k [tex]\frac{q1 *q2}{ r^{2} }[/tex]

Where q1, q2= magnitude of the charge in C.

and r = the distance between them in m.

k= constant of proportionality.

Question given data,

F1=18N

let r=x (say)

then F1=18=k [tex]\frac{q1 *q2}{ x^{2} }[/tex].....................(1)

if r is halved i.e r=x/2

now new force

F2=k[tex]\frac{q1 *q2}{ (x/2)^{2} }[/tex]

F2=4×k[tex]\frac{q1 *q2}{ x^{2} }[/tex]

F2=4×F1....................(∵F1=k [tex]\frac{q1 *q2}{ x^{2} }[/tex])

=4×18..............................(∵F1=18)

F2=72N

Hence The force between two charged particles when the distance between them is halved is 72N.

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NASA's Juno spacecraft has been orbiting Jupiter since 2016, providing scientists with unprecedented views of the gas giant. One of the most striking features of Jupiter is its Great Red Spot, a massive storm that has been raging for centuries.

What is Great Red Spot?

On Jupiter, a huge storm known as the Big Red Spot has been raging for at least 400 years and possibly longer. It is a zone of persistently high pressure that is present in the planet's southern hemisphere. The storm, which is thought to be around two to three times the size of our globe, is so enormous that it could easily devour the entire planet.

When it began orbiting Jupiter in 2016, NASA's Juno probe has given scientists hitherto unheard-of pictures of the gas giant. The Great Red Spot, a huge storm that has been raging for ages, is one of Jupiter's most stunning characteristics. The polar regions of Jupiter are however home to numerous additional storms, including sizable and intricate cyclones.

The scale and intricacy of the cyclones on Jupiter's poles, especially a centre cyclone on the north pole known as "Jupiter's North Pole Hexagon," first astounded scientists. These storms can endure for decades or even centuries and are more larger and better organised than any cyclones on Earth.

It's not apparent to me which tornado you're referring about in particular. But, researchers looking at Jupiter's The variety and complexity of these storms have probably fascinated storms, and they may be attempting to comprehend the underlying physical mechanisms. Jupiter's storms can be studied to learn more about planetary weather, including our own.

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Explanation:

E1 = k Q1 / (0.075 m)^2

= 9 x 10^9 Nm^2/C^2 * 10 x 10^-9 C / 0.075^2 m^2

= 400 N/C

E2 = k Q2 / (0.075 m)^2

= 9 x 10^9 Nm^2/C^2 * (-3 x 10^-9 C) / 0.075^2 m^2

= -120 N/C

Since the electric fields due to the two charges are in opposite directions, we can find the net electric field at the midway point by taking the difference between the two:

E = E1 - E2

= 400 N/C - (-120 N/C)

= 520 N/C

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings.

dQ/dt = -k(T - Troom)

where Q is the amount of heat in the water, t is time, k is a constant, T is the temperature of the water, and Troom is the room temperature.

Using the given information, we can set up the following system of equations:

100 = k(100 - 21) (at t = 0) 65 = k(100 - 21) e⁽⁻¹⁰⁾k (at t = 10)

Solving for k in the first equation gives k = 1/79. Plugging this value into the second equation and solving for T gives T = 36.1 degrees Celsius.

Now, to find the temperature after an additional 5 minutes, we can use the same equation with a new time value:

T = (65 - Troom) e^(-k(10+5)) + Troom

Plugging in the values we know gives T = 30.4 degrees Celsius.

Therefore, the temperature of the water after an additional 5 minutes of cooling is approximately 30.4 degrees Celsius

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At wavelengths [tex]$644.44 \mathrm{~nm}$[/tex] and[tex]$483.33 \mathrm{~nm}$[/tex], the stars [tex]$\mathrm{A}$[/tex] and B have maximum intensities. temperature of most intense star is 6,502.242K

Temperature of star [tex]$\mathrm{A}$[/tex] is [tex]$\mathrm{TA}[/tex]=4,500 [tex]\mathrm{~K}$.[/tex]

Temperature of star $B$ is TB $=6,000 [tex]\mathrm{~K}$.[/tex]

Wavelength of most intense star is [tex]$\lambda_0=446[/tex]

We have to find wavelengths corresponding to maximum intensity of stars at temperatures [tex]$T \wedge$[/tex]and TB and wavelength of most intense star.

From Wein's law, we have

[tex]$\lambda_{\mathrm{nm}}=\frac{2.90 \times 106}{\mathrm{Tk}_{\mathrm{k}}}$[/tex]

Where [tex]$\lambda_{\mathrm{nm}}$[/tex] is in nanometers and [tex]$\mathrm{T} k$[/tex] is in Kelvin.

For Star A, equation (1) becomes

[tex]$\lambda_{\mathrm{A}}=\frac{2.90 \times 106}{\mathrm{TA}}$[/tex]

Using values in (2), we get

[tex]\lambda_{\mathrm{A}}[/tex] & =[tex]\frac{2.90 \times 106}{4,500 \mathrm{~K}} \\[/tex]& =644.44 [tex]\mathrm{~nm}[/tex]

For Star B equation (1) becomes

[tex]$$\lambda_{\mathrm{B}}=\frac{2.90 \times 106}{\mathrm{Tr}}$$[/tex]

Let the temperature of the most intense star is[tex]$\mathrm{T} 0$.[/tex] Now, equation (1) for most intense star becomes [tex]$\lambda_0=\frac{2.90 \times 106}{T_0}$[/tex]

From above equation, we have

[tex]$\mathrm{T}_0=\frac{2.90 \times 106}{\lambda_0}$[/tex]

Using values in (3), we get

[tex]$$\begin{aligned}\lambda_{\mathrm{B}} & =\frac{2.90 \times 106}{6,000 \mathrm{~K}} \\& =483.33 \mathrm{~nm}\end{aligned}$$[/tex]

Using vales in (4), we get

[tex]\begin{aligned}\mathrm{T}_0 & =\frac{2.90 \times 10_6}{446 \mathrm{~nm}} \\= & 6,502.242 \mathrm{~K}\end{aligned}[/tex]

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